In circle T below, find the measure of angle BEC (This is actually Geometry so..)

Explain why the expression 10 to the power of zero, one to the power of four and 1×1.0 to the power of zero have the same value

Nina [5.8K]

Answer:

10*0=1, according to the law of indices which states that anything to the zero power is equal to one

1*4=1×1×1×1=1

1×1.0=1.0=1

3 0

3 years ago

JUST ANSWER PLEASE!!! QUICK

jeka57 [31]

Answer:

Options 1 and 3

Step-by-step explanation:

We should know that, the system of linear equations can be treated as matrices, i.e: we can modify or make any operations provide that we must apply the same operation for all terms of each equation.

Given: the solution for the following system is (2,9)

Px + Qy = R ⇒(1)

Tx + Uy = V ⇒(2)

We will check which system of equation has the same solution.

System A) Px + Qy = R

(P+T)x + (Q+U)y = R+V ⇒(3)

So, By summing (1) and (2) we will get the equation (3)

So, system A has the same solution (2,9)

System B) Px + Qy = R

(P+2T)x + (Q+2U)y = R-2V ⇒(4)

By multiplying equation (2) by 2 and add with equation (1), we will get:

(P+2T)x + (Q+2U)y = R+2V

Which is not the same as equation (4)

So, system B has not the same solution (2,9)

System C) (T-P)x + (U-Q)y = V-R ⇒(5)

Tx + Uy = V

By multiplying equation (1) by -1 and add with equation (2), we will get the equation (5)

So, system C has the same solution of (2,9)

System D) (T-P)x + (Q+U)y = V-R ⇒(6)

Tx + Uy = V

We cannot get equation (6) by the same operation over equation (1)

Note the coefficient of x and y⇒ (T-P) and (Q+U)

They must be (T+P) and (Q+U) OR (T-P) and (Q-U)

So, system D has not the same solution of (2,9)

System E) (5T-P)x + (5U-Q)y = V-5R ⇒ (6)

Tx + Uy = V

By subtract equation (1) from 5 times equation (2), we will get:

(5T-P)x + (5U-Q)y = 5V-R

Which is not the same as equation (6)

So, system E has not the same solution (2,9)

As a conclusion, the systems which have the same solution are:

Options 1 and 3

5 0

2 years ago

The diameter of a particle of contamination (in micrometers) is modeled with the probability density function f(x)= 2/x^3 for x

natulia [17]

Answer:

a) 0.96

b) 0.016

c) 0.018

d) 0.982

e) x = 2

Step-by-step explanation:

We are given with the Probability density function f(x)= 2/x^3 where x > 1.

Firstly we will calculate the general probability that of P(a < X < b)

P(a < X < b) = $\int_{a}^{b} \frac{2}{x^{3}} dx$ = $2\int_{a}^{b} x^{-3} dx$

= $2[ \frac{x^{-3+1} }{-3+1}]^{b}_a dx$ { Because $\int_{a}^{b} x^{n} dx = [ \frac{x^{n+1} }{n+1}]^{b}_a$ }

= $2[ \frac{x^{-2} }{-2}]^{b}_a$ = $\frac{2}{-2} [ x^{-2} ]^{b}_a$

= $-1 [ b^{-2} - a^{-2} ]$ = $\frac{1}{a^{2} } - \frac{1}{b^{2} }$

a) Now P(X < 5) = P(1 < X < 5) {because x > 1 }

Comparing with general probability we get,

P(1 < X < 5) = $\frac{1}{1^{2} } - \frac{1}{5^{2} }$ = $1 - \frac{1}{25}$ = 0.96 .

b) P(X > 8) = P(8 < X < ∞) = 1/ $8^{2}$ - 1/∞ = 1/64 - 0 = 0.016

c) P(6 < X < 10) = $\frac{1}{6^{2} } - \frac{1}{10^{2} }$ = $\frac{1}{36} - \frac{1}{100 }$ = 0.018 .

d) P(x < 6 or X > 10) = P(1 < X < 6) + P(10 < X < ∞)

= $(\frac{1}{1^{2} } - \frac{1}{6^{2} })$ + (1/ $10^{2}$ - 1/∞) = 1 - 1/36 + 1/100 + 0 = 0.982

e) We have to find x such that P(X < x) = 0.75 ;

⇒ P(1 < X < x) = 0.75

⇒ $\frac{1}{1^{2} } - \frac{1}{x^{2} }$ = 0.75

⇒ $\frac{1} {x^{2} }$ = 1 - 0.75 = 0.25

⇒ $x^{2}$ = $\frac{1}{0.25}$ ⇒ $x^{2}$ = 4 ⇒ x = $2$

Therefore, value of x such that P(X < x) = 0.75 is 2.

8 0

3 years ago

A. Evaluate the polynomial

sdas [7]

Answer:

a) y = x³− 5x² + 6x + 0.55 at x = 1.37.

Use 3-digit arithmetic with chopping. Evaluate the percent relative round-off error.

-1.183%

b. Express y as y = ((x − 5)x + 6)x + 0.55 (this is the same equation). Use again 3-digit arithmetic with chopping. Evaluate the percent relative round-off error and compare with part (a). Make the conclusion about which form of the polynomial is superior.

-0.161%

Comparing part a and b together, part b is more superior because the percent(%) error is smaller when compared to part a

Step-by-step explanation:

a) y = x³− 5x² + 6x + 0.55 at x = 1.37.

Use 3-digit arithmetic with chopping. Evaluate the percent relative round-off error.

Let's evaluate before applying the 3 digit arithmetic chopping rule

y = 1.37³ - 5 × 1.37² + 6 × 1.37 + 0.55

y = 1.956853

Let evaluate each components of the polynomial one by one

Note that: 3-digit arithmetic chopping means to approximate chop off or remove number after the 3 significant figures.

y = x³− 5x² + 6x + 0.55 at x = 1.37.

x³ = 1.37³ = 2.571353

≈ 2.57

x² = 1.37² = 1.8769

≈ 1.88

5x² = 1.87 × 5

= 9.35

x = 1.37

6x = 1.37 × 6

6x = 8.22

Evaluating the polynomial

y = 2.57 - 9.38 + 8.22 + 0.55

y = 1.98

The percent relative round-off error =

1.956853 - 1.98/1.956853 × 100

= -1.183%

b. Express y as y = ((x − 5)x + 6)x + 0.55 (this is the same equation). Use again 3-digit arithmetic with chopping. Evaluate the percent relative round-off error and compare with part (a). Make the conclusion about which form of the polynomial is superior.

y = ((x − 5)x + 6)x + 0.55

Evaluating with the 3 digit chop off rule is applied

= ((1.37 - 5)1.37 + 6)1.37 + 0.55

=( 1.8769 - 6.85) + 6) 1.37 + 0.55

= (- 4.9731 + 6 )1.37 + 0.55

= 1.0269 × 1.37 + 0.55

= 1.406853

= 1.956853.

≈ 1.96

Note in: evaluating before applying the 3 digit arithmetic chopping rule

y = 1.37³ - 5 × 1.37² + 6 × 1.37 + 0.55

y = 1.956853

The percent relative round-off error

1.956853 - 1.96/1.956853 × 100

= -0.161%

Comparing part a and b together, part b is more superior because the percent(%) error is smaller when compared to part a

3 0

3 years ago

Y=1/2x+2 y=1/3x+2 solve for graphing

madreJ [45]

Answer:

I plotted it out on the graph and here it is:

6 0

2 years ago